Skip to main content
Search type Search syntax
Tags [tag]
Exact "words here"
Author user:1234
user:me (yours)
Score score:3 (3+)
score:0 (none)
Answers answers:3 (3+)
answers:0 (none)
isaccepted:yes
hasaccepted:no
inquestion:1234
Views views:250
Code code:"if (foo != bar)"
Sections title:apples
body:"apples oranges"
URL url:"*.example.com"
Saves in:saves
Status closed:yes
duplicate:no
migrated:no
wiki:no
Types is:question
is:answer
Exclude -[tag]
-apples
For more details on advanced search visit our help page
Results tagged with
Search options not deleted user 12978

Banach spaces, function spaces, real functions, integral transforms, theory of distributions, measure theory.

13 votes
1 answer
3k views

Does this metric have an official name? Lévy metric? Ky Fan metric?

Let $X$ and $Y$ be random variables taking values in a separable metric space $(S,d)$. The metric I have in mind is $$\rho(X,Y) = \mathbb{E}[\min\{d(X,Y),1\}]$$ if $X$ and $Y$ take values in the a me …
5 votes
2 answers
639 views

Is the Hausdorff metric on sub-$\sigma$-fields separable?

Let $(X,\mu,\mathcal{F})$ be a probability space. The paper Equiconvergence of Martingales by Edward Boylan introduced a pseudometric on sub-$\sigma$-fields (sub-$\sigma$-algebras) of $\mathcal{F}$ a …
9 votes
2 answers
535 views

What mode of convergence is this?

I'm interested in a new (to me) mode of convergence which is stronger than convergence in measure/probability. I want to know if it has a name and if it is used much in the literature. I will write …
15 votes
2 answers
3k views

Generalizations of the Tietze extension theorem (and Lusin's theorem)

I am reasking a year-old math.stackexchange.com question asked by someone else. (For my needs every space $X$ and $Y$ will be Polish---that is a completely separably metrizable space.) The Tietze ex …
8 votes
0 answers
382 views

What is the name for a Banach space property closed under ultraproducts?

In Banach space theory, a super-property is a property of a Banach space that is preserved under ultrapowers. (Update (2015-09-28): The property must also be closed under isometric embeddings.) (Sup …
4 votes
2 answers
505 views

Martingale-cotype vs cotype on super-reflexive spaces

I'm have difficultly nailing down the direction of some implications. For $2 \leq q < \infty$, there are (at least) two ways to say that a Banach space $B$ has "cotype $q$". $B$ has cotype q. $B$ is …
2 votes
0 answers
169 views

Classify spaces that make extension theorems hold

Recall a Polish space is a completely metrizable separable space. Say a Polish space $Y$ is a terminal space if for any Polish space $X$ and any closed $C \subseteq X$, one can extend a continuous m …
4 votes
0 answers
309 views

Conditional expectation with respect to random closed sets

Short question If $X$ is a random closed set, and $Y$ is an integrable random variable, I would like a definition of the "conditional expectation" $$\mathbf{E}[Y \mid X\ni x].$$ Has this been worked …
5 votes
2 answers
434 views

Sz.-Nagy dilation for uniformly convex Banach spaces

The Sz.-Nagy dilation theorem says that for a Hilbert space $H$ with nonexpansive operator $T$, there is a larger space $H'$ containing $H$ and a unitary operator $U$ on $H'$ such that for all $x \in …
3 votes
Accepted

Martingale-cotype vs cotype on super-reflexive spaces

I'll attempt to answer this with what I've found: (1) The answer to my main question is that it is not true. In this document of Pisier's, he states It is possible to find a uniformly convex s …
Jason Rute's user avatar
  • 6,287
1 vote
1 answer
171 views

Extension of power bounded operators over a finite subspace

Suppose $Y$ is a Banach space and $X$ is a finite-dimensional subspace of $Y$. Further assume $T:X \rightarrow X$ is a linear operator which is power bounded from above and below, in other words ther …